First slide

Gravitaional potential energy

Question

A body is projected vertically upward from the surface of the earth with a velocity equal to half the escape velocity. If fi is the radius of the earth, the maximum height attained by the body is

Difficult

A

R/3
B

R /6
C

2R/3
D

R

Solution

initial K.E. $= \frac{1}{2} {mv}^{2} = \frac{1}{2} m {(\frac{v_{e}}{2})}^{2}$
$= \frac{1}{8} {mv}_{e}^{2} = \frac{GmM}{4 R} [∵ v_{e} = \sqrt{(\frac{2 MG}{R})}]$
$P . E . = - \frac{GmM}{R}$
Total Energy $= \frac{GmM}{4 R} - \frac{GmM}{R}$
$= - \frac{3 GmM}{4 R}$ ……………….(1)
Lot the body come to rest at a height r from the centre of the earth. Now its energy is given by
Final energy $= - \frac{GmM}{r}$ ………..(2)
Comparing eqs. (1) and (2), we get
$r = (4 R / 3)$
$∴$ Maximum height attained
$= r - R = (4 R / 3) - R = (R / 3)$

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